Theorem – If $K$ is a field, $\overline{K}$ its algebraic closure, and $I$ a proper ideal of $K[X_1,\cdots,X_n]$, then $\mathcal{V}(I) \neq \emptyset$ in $\overline{K}^n$.
This is the weak version of Hilbert's Nullstellensatz theorem, but it is not considering an algebraically closed field. I'm having trouble adapting this version.
In the classic case, when the field is algebraically closed, we use a result that says who are the maximal ideals in $K[X_1,\cdots,X_n]$. However, since in this case the field is not algebraically closed, I'm having a hard time. I really appreciate the help. I really need it.
Best Answer
Clearly it's enough to prove the assertion when $ I $ is maximal so let us assume that it is. In that case, Zariski's lemma implies that $ k[X_1, \cdots , X_n]/I $ is a finite field extension of $ k $, call it $ L $. Then $$ \bar{k} [X_1, \cdots, X_n] / I = k[X_1, \cdots , X_n]/I \otimes_k \bar{k} = L \otimes \bar{k} $$ The last object is now a finitely generated $ \bar{k} $-algebra of finite dimension $ [L:k] \ge 1 $ over $ \bar{k} $ as a $ \bar{k} $-vector space, in particular a non-zero ring, hence it has a maximal ideal. That is $ I \subset \bar{k} [X_1, \cdots , X_n ] $ is contained in a maximal ideal, which is, say $ (X_1 - a_1 , \cdots, X_n - a_n) $ by the usual Nullstellensatz. Then $ (a_1, \cdots , a_n) $ is in $ V(I) $ in $ \bar{k}^n $ finishing the proof.